1L of a buffer composed of acetic acid and sodium acetate has a pH of 4.3. Adding 10mL of 2M sodium hydroxide solution to 100mL of this buffer causes the pH to rise to 4.87. what is the total molarity of the original buffer?

Answer 1

pH = pka + log b/a
4.3 = 4.74 + log b/a
b/a = 0.363
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add 100 mL x 2M = 20 mmols to obtain
4.87 = 4.74 + log b/a
b/a = 1.35 and redo to read
1.349 = (b+20)/(a-20)
1.349(a-20) = b+20
1.349a - 27 = b+20
Substitute 0.363 for b to obtain
1.349a - 27 = 0.363a + 20
0.986a = 47
a = 47.7
b = 0.363a = 0.363*47.7 = 17.3
a + b = 47.7 + 17.3 = 65
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Check:
..........HAc + OH^- ==> Ac^- + H2O
I.........47.7...0.......17.3
add ............20.........
C.........-20..-20.......+20
E.........27.7...0.......37.3

pH = 4.74 + log(37.3/27.7)
pH = 4.87

Answer 2

i got a total concentration of 0.0774M is this correct?

Answer 3

Where are you getting that?